Question

Find the voltage of power supply to which a bulb of rating 200W- 220V is connected to consume energy at the rate of 100 J/s.yug

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Answer 1

Answer:

In this case, Power P = 100 W and Voltage V = 220 V. So, current I is I =

P/V = 100/220

= 0.45 A. When the supplied voltage is 110 V, the power consumed will be

P=VI = 110 * 0.45 = 50 W. Hence, the power consumed when operated at 110 V is 50 W.

Step-by-step explanation:

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Answer 2

Answer:

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Step-by-step explanation:

Resistance R=V

Resistance R=V 2

Resistance R=V 2 /P=48400/200=242Ω

Resistance R=V 2 /P=48400/200=242ΩI

Resistance R=V 2 /P=48400/200=242ΩI rms

Resistance R=V 2 /P=48400/200=242ΩI rms

Resistance R=V 2 /P=48400/200=242ΩI rms =V

Resistance R=V 2 /P=48400/200=242ΩI rms =V rms

Resistance R=V 2 /P=48400/200=242ΩI rms =V rms

Resistance R=V 2 /P=48400/200=242ΩI rms =V rms /R=220/242=0.909A

Resistance R=V 2 /P=48400/200=242ΩI rms =V rms /R=220/242=0.909APeak Value of current =I

Resistance R=V 2 /P=48400/200=242ΩI rms =V rms /R=220/242=0.909APeak Value of current =I rms

Resistance R=V 2 /P=48400/200=242ΩI rms =V rms /R=220/242=0.909APeak Value of current =I rms

Resistance R=V 2 /P=48400/200=242ΩI rms =V rms /R=220/242=0.909APeak Value of current =I rms ×

Resistance R=V 2 /P=48400/200=242ΩI rms =V rms /R=220/242=0.909APeak Value of current =I rms × 2

Resistance R=V 2 /P=48400/200=242ΩI rms =V rms /R=220/242=0.909APeak Value of current =I rms × 2

Resistance R=V 2 /P=48400/200=242ΩI rms =V rms /R=220/242=0.909APeak Value of current =I rms × 2 2

Resistance R=V 2 /P=48400/200=242ΩI rms =V rms /R=220/242=0.909APeak Value of current =I rms × 2 2

Resistance R=V 2 /P=48400/200=242ΩI rms =V rms /R=220/242=0.909APeak Value of current =I rms × 2 2 =0.909×1.414=1.285A

Resistance R=V 2 /P=48400/200=242ΩI rms =V rms /R=220/242=0.909APeak Value of current =I rms × 2 2 =0.909×1.414=1.285A (2)

Resistance R=V 2 /P=48400/200=242ΩI rms =V rms /R=220/242=0.909APeak Value of current =I rms × 2 2 =0.909×1.414=1.285A (2)Power P=V

Resistance R=V 2 /P=48400/200=242ΩI rms =V rms /R=220/242=0.909APeak Value of current =I rms × 2 2 =0.909×1.414=1.285A (2)Power P=V 1

Resistance R=V 2 /P=48400/200=242ΩI rms =V rms /R=220/242=0.909APeak Value of current =I rms × 2 2 =0.909×1.414=1.285A (2)Power P=V 12

Resistance R=V 2 /P=48400/200=242ΩI rms =V rms /R=220/242=0.909APeak Value of current =I rms × 2 2 =0.909×1.414=1.285A (2)Power P=V 12

Resistance R=V 2 /P=48400/200=242ΩI rms =V rms /R=220/242=0.909APeak Value of current =I rms × 2 2 =0.909×1.414=1.285A (2)Power P=V 12 /R=200×200/242=165.28W